Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following:

(A) $(7x + 4y)^2$                       (B) $\left(\frac{7x}{5} + \frac{3y}{2}\right)^2$

(C) $(2.5p + 1.5q)^2$              (D) $\left(\frac{3s}{4} + 8t\right)^2$

(E) $\left(x + \frac{1}{2y}\right)^2$                         (F) $\left(\frac{1}{x} + \frac{1}{y}\right)^2$                                                  [Page No. 71]

(A) $(7x)^2 + 2(7x)(4y) + (4y)^2$

$= 49x^2 + 56xy + 16y^2$

(B) $\left(\frac{7x}{5}\right)^2 + 2\left(\frac{7x}{5}\right)\left(\frac{3y}{2}\right) + \left(\frac{3y}{2}\right)^2$

$= \frac{49x^2}{25} + \frac{21xy}{5} + \frac{9y^2}{4}$

(C) $(2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$

$= 6.25p^2 + 7.5pq + 2.25q^2$

(D) $\left(\frac{3s}{4}\right)^2 + 2\left(\frac{3s}{4}\right)(8t) + (8t)^2$

$= \frac{9s^2}{16} + 12st + 64t^2$

(E) $x^2 + 2x\left(\frac{1}{2y}\right) + \left(\frac{1}{2y}\right)^2 = x^2 + \frac{x}{y} + \frac{1}{(4y^2)}$

$$\left( \frac{1}{x} \right)^2 + 2 \left( \frac{1}{x} \right) \left( \frac{1}{y} \right) + \left( \frac{1}{y} \right)^2 = \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}$$

Caution

Students should remember that $(a + b)^2 = a^2 + 2ab + b^2$, and not $a^2 + b^2$. The middle term $2ab$ must never be left out.